5xy - 5x + y = 5

<=> 5xy = 5 + 5x - y

<=> \(\left\{{}\begin{matrix}x=\dfrac{5+5x-y}{5y}\\y=\dfrac{5+5x-y}{5x}\end{matrix}\right.\)

\(5xy-5x+y=5\)

\(\Rightarrow5x\left(y-1\right)+\left(y-1\right)=4\)

\(\Rightarrow\left(y-1\right)\left(5x+1\right)=4\)

Do \(x,y\in Z\)

TH1: \(\left\{{}\begin{matrix}y-1=1\\5x+1=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}y-1=4\\5x+1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=5\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

TH3: \(\left\{{}\begin{matrix}y-1=2\\5x+1=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=\dfrac{1}{5}\left(ktm\right)\end{matrix}\right.\)

TH4: \(\left\{{}\begin{matrix}y-1=-2\\5x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\left(tm\right)\\x=-\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\)

TH5: \(\left\{{}\begin{matrix}y-1=-1\\5x+1=-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

TH6: \(\left\{{}\begin{matrix}y-1=-4\\5x+1=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\left(tm\right)\\x=-\dfrac{2}{5}\left(ktm\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(0;5\right);\left(-1;0\right)\right\}\)

5xy-5x+y=5

 suy ra 5xy-5x+y-1=6

 suy ra 5x(y-1)+(y-1)=6

 suy ra (5x+1)(y-1)=6

x,y thuôc Z  suy ra 5x+1,y-1 thuộc Z

 suy ra 5x+1,y-1 thuộc Ư(6)

Ta có bảng sau:

5xy(5x+y = 5 

ta có 5 = 5 x 1 = 1 x5

ta xét 2 TH 

TH1 5xy = 5 => xy = 1

và 5x +y = 1 => y = 1 - 5x 

từ đây tự giải ra 

và làm tiếp tH 2 nha

x; y thuộc Z

(x;y)=(-1;0)

Bạn cho mình biết vì sao với

a)x=7;y=5

b)x=0;y=5

-5xy-5x+y=5 <=> -5x(y+1) +y+1 = 6 <=> (y+1)(1-5y) =6

ok.. đến đây bạn tự giải tiếp.. tick cho mik nha. :v

\(\text{a) }\left(5x+1\right)\left(y-1\right)=4\)

\(\Leftrightarrow5x+1,y-1\inƯ\left(4\right)\)

\(\Leftrightarrow5x+1,y-1\in\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng :

\(\text{b) }5xy-5x+y=5\)

\(\Leftrightarrow\left(5xy+y\right)-5x=5\)

\(\Leftrightarrow y\left(5x+1\right)-\left(5x+1\right)-1=5-1\)

\(\Leftrightarrow y\left(5x+1\right)-\left(5x+1\right)-1=4\)

\(\Leftrightarrow\left(y-1\right).\left(5x+1\right)=4\)

\(\Leftrightarrow y-1,5x+1\inƯ\left(4\right)\)

\(\Leftrightarrow y-1,5x+1\in\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng :

2xy - 8x - y = 17

=> 2x[y - 1] - y = 17

=> 2x[y - 1] - y + 1= 18

=> 2x[y - 1] - [y - 1] = 18

=> [2x - 1][y-1] = 18

Mà 2x - 1 lẻ nên 2x - 1 \(\in\left\{-9;-3;-1;1;3;9\right\}\)

Ta có:

Vậy; .........

5xy - 5x + y = 5

=> 5x[y - 1] + y = 5

=> 5x[y-1] + y - 1 = 4

=> 5x[y-1] + [y-1] = 4

=> [5x - 1][y-1] = 4

Ta có:

Vậy:.........

2xy + x - 6y = 14

=> 2xy - 6y + x = 14

=> 2y[x - 3] + x = 14

=> 2y[x-3] + x-3 = 11

=> 2y[x-3] + [x-3] = 11

=> [2y - 1][x-3] = 11

Ta có:

Vậy:...........